What is the integral of 9tan^3(2x)sec^5(2x) ?

The integral of 9tan^3(2x)sec^5(2x) is 9((sec^7(2x))/(14)-(sec^5(2x))/(10))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9tan^3(2x)sec^5(2x)

\int 9 tan^{3} (2 x) sec^{5} (2 x) d x

9 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}) + C

Solution steps

\int 9 tan^{3} (2 x) sec^{5} (2 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int tan^{3} (2 x) sec^{5} (2 x) d x

Rewrite using trig identities

= 9 \cdot \int (- 1 + sec^{2} (2 x)) tan (2 x) sec^{5} (2 x) d x

Apply u-substitution

: \int \frac{u ^{4} ( u ^{2} - 1 )}{2} d u

= 9 \cdot \int \frac{u ^{4} ( u ^{2} - 1 )}{2} d u

Expand

\frac{u ^{4} ( u ^{2} - 1 )}{2} : \frac{u ^{6}}{2} - \frac{u ^{4}}{2}

= 9 \cdot \int \frac{u ^{6}}{2} - \frac{u ^{4}}{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (\int \frac{u ^{6}}{2} d u - \int \frac{u ^{4}}{2} d u)

\int \frac{u ^{6}}{2} d u = \frac{u ^{7}}{14}

\int \frac{u ^{4}}{2} d u = \frac{u ^{5}}{10}

= 9 (\frac{u ^{7}}{14} - \frac{u ^{5}}{10})

Substitute back

u = sec (2 x)

= 9 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10})

Add a constant to the solution

= 9 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}) + C

\int_{- 5}^{5} 1 + x^{2} d x

(e^{y})^{^{'}}

\int \frac{1}{( x ^{2} - 36 ) ^{\frac{3}{2}}} d x

\int \frac{1 - x ^{2}}{2} d x

csc^{2} (x) - 5

What is the integral of 9tan^3(2x)sec^5(2x) ?
The integral of 9tan^3(2x)sec^5(2x) is 9((sec^7(2x))/(14)-(sec^5(2x))/(10))+C